Chapter SH, Problem 9.14EP

To determine

To find: the equation of the hyperbola shown in the graph.

Answer to Problem 9.14EP

The equation of the hyper bola her is $\text{175}{\left(\text{x+1}\right)}^{\text{2}}-\text{81}{\left(\text{y+3}\right)}^{\text{2}}\text{=14175}$ .

Explanation of Solution

Given:

  

Concept used:

The equation of hyper having foci at horizontal axis is:

  $\frac{{\left(\text{x-h}\right)}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{-}\frac{{\left(\text{y-k}\right)}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1}$ .

  ${\text{c}}^{\text{2}}{\text{=a}}^{\text{2}}{\text{+b}}^{\text{2}}$ .

Vertices is denoted by $\text{a}$ .

Co-vertices is denoted by $\text{b}$ .

Foci is denoted by $\text{c}$ .

Calculation:

The equation of hyper having foci at horizontal axis is:

  $\frac{{\left(\text{x-h}\right)}^{\text{2}}}{{\text{a}}^{\text{2}}}\text{-}\frac{{\left(\text{y-k}\right)}^{\text{2}}}{{\text{b}}^{\text{2}}}\text{=1}$ ........... $\left(\text{i}\right)$

Asymptotes will cross at $\left(-1,-3\right)$ .

So, the center= $\left(\text{h,k}\right)=\left(-1,-3\right)$ …… $\left(\text{ii}\right)$

Foci $\text{=}\left(\text{0,c}\right)\text{,}\left(\text{0,-c}\right)=\left(0,5\right),\left(0,-5\right)$ .

Vertices= $\text{a}$

Here from the given graph:

  $\text{2a=10-}\left(-8\right)$ .

  $\Rightarrow \text{a=}\frac{18}{2}=9$ .

  $\text{a}=9.$

foci $\text{=c}$ .

Here from the given graph:

  $\text{2c=16-}\left(\text{-16}\right)$ .

  $\Rightarrow \text{c=}\frac{\text{32}}{2}=16$ .

  $\text{c}=16.$

Co- vertices $\text{=}\left(\text{0,b}\right)\text{,}\left(\text{0,-b}\right)$ .

Co-vertices can be calculated as:

  ${\text{c}}^{\text{2}}{\text{=a}}^{\text{2}}{\text{+b}}^{\text{2}}$ .

  $\Rightarrow {16}^{\text{2}}{\text{=9}}^{\text{2}}{\text{+b}}^{\text{2}}$ .

  $\Rightarrow {\text{256-81=b}}^{\text{2}}$ .

  ${\text{b}}^{\text{2}}=175$ …… $\left(\text{iii}\right)$

Similarly, ${\text{a}}^2=81$ …. $\left(\text{iv}\right)$

Putting equation $\left(\text{ii}\right)$ , $\left(\text{iii}\right)$ and $\left(\text{iv}\right)$ in equation $\left(\text{i}\right)$ :

  $\frac{{\left(\text{x-}\left(\text{-1}\right)\right)}^{\text{2}}}{\text{81}}\text{-}\frac{{\left(\text{y-}\left(\text{-3}\right)\right)}^{\text{2}}}{\text{175}}\text{=1}$ .

  $\frac{{\left(\text{x+1}\right)}^{\text{2}}}{\text{81}}\text{-}\frac{{\left(\text{y+3}\right)}^{\text{2}}}{\text{175}}\text{=1}$ .

  $\frac{\text{175}{\left(\text{x+1}\right)}^{\text{2}}-\text{81}{\left(\text{y+3}\right)}^{\text{2}}}{\text{81}\times \text{175}}\text{=1}$ .

  $\text{175}{\left(\text{x+1}\right)}^{\text{2}}-\text{81}{\left(\text{y+3}\right)}^{\text{2}}\text{=14175}$ .

Hence, the equation of the hyper bola her is $\text{175}{\left(\text{x+1}\right)}^{\text{2}}-\text{81}{\left(\text{y+3}\right)}^{\text{2}}\text{=14175}$ .